Diode reverse recovery numerical (3 µs, 30 A/µs)

Reverse recovery: when a conducting diode is switched to reverse bias, its current does not stop at zero. The minority carriers stored in the junction must first be removed, so the current falls through zero, flows in the reverse direction for a short time and then decays to zero.

  iD
   ^ IF
   |-------.
   |        \  slope = di/dt
   |         \
 --+----------\--------------------> t
   |           \      .--------
   |            \   .'
   |   - IRR ....\.'   tb
   |          ta  |<->|
   |         |<-->|
   |         |<---trr--->|

• $t_a$: time from current zero to the peak reverse current $I_{RR}$, while charge stored in the depletion region is removed.
• $t_b$: time for the reverse current to decay from $I_{RR}$ to about $0.25 I_{RR}$, while charge in the bulk is removed.
• Reverse recovery time $t_{rr} = t_a + t_b$; softness factor $S = t_b / t_a$.
• Peak reverse current: $I_{RR} = t_a \dfrac{di}{dt}$.
• Reverse recovery (stored) charge $Q_{RR}$ is the area under the reverse current, approximately a triangle:

$$Q_{RR} = \tfrac{1}{2} I_{RR}\, t_{rr}$$

Numerical

Given: $t_{rr} = 3\ \mu$s, $di/dt = 30$ A/µs. No softness factor is given, so the usual assumption of abrupt recovery ($t_b \ll t_a$, so $t_{rr} \approx t_a$) is used.

$$\begin{aligned} Q_{RR} &= \tfrac{1}{2}\frac{di}{dt}\, t_{rr}^2 = \tfrac{1}{2} \times 30 \times 10^{6} \times (3 \times 10^{-6})^2 \\ &= 135 \times 10^{-6}\ \text{C} = 135\ \mu\text{C} \\ I_{RR} &= \sqrt{2 Q_{RR}\frac{di}{dt}} = \sqrt{2 \times 135 \times 10^{-6} \times 30 \times 10^{6}} \\ &= \sqrt{8100} = 90\ \text{A} \end{aligned}$$

(Check: $I_{RR} = t_{rr}\,di/dt = 3 \times 30 = 90$ A.)

Answer: stored charge $Q_{RR} = 135\ \mu$C, peak reverse current $I_{RR} = 90$ A.